diff --git a/leetcode/2182. Construct String With Repeat Limit/README.md b/leetcode/2182. Construct String With Repeat Limit/README.md
new file mode 100644
index 00000000..d7c10524
--- /dev/null
+++ b/leetcode/2182. Construct String With Repeat Limit/README.md	
@@ -0,0 +1,87 @@
+# [2182. Construct String With Repeat Limit (Medium)](https://leetcode.com/problems/construct-string-with-repeat-limit/)
+
+<p>You are given a string <code>s</code> and an integer <code>repeatLimit</code>. Construct a new string <code>repeatLimitedString</code> using the characters of <code>s</code> such that no letter appears <strong>more than</strong> <code>repeatLimit</code> times <strong>in a row</strong>. You do <strong>not</strong> have to use all characters from <code>s</code>.</p>
+
+<p>Return <em>the <strong>lexicographically largest</strong> </em><code>repeatLimitedString</code> <em>possible</em>.</p>
+
+<p>A string <code>a</code> is <strong>lexicographically larger</strong> than a string <code>b</code> if in the first position where <code>a</code> and <code>b</code> differ, string <code>a</code> has a letter that appears later in the alphabet than the corresponding letter in <code>b</code>. If the first <code>min(a.length, b.length)</code> characters do not differ, then the longer string is the lexicographically larger one.</p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+
+<pre><strong>Input:</strong> s = "cczazcc", repeatLimit = 3
+<strong>Output:</strong> "zzcccac"
+<strong>Explanation:</strong> We use all of the characters from s to construct the repeatLimitedString "zzcccac".
+The letter 'a' appears at most 1 time in a row.
+The letter 'c' appears at most 3 times in a row.
+The letter 'z' appears at most 2 times in a row.
+Hence, no letter appears more than repeatLimit times in a row and the string is a valid repeatLimitedString.
+The string is the lexicographically largest repeatLimitedString possible so we return "zzcccac".
+Note that the string "zzcccca" is lexicographically larger but the letter 'c' appears more than 3 times in a row, so it is not a valid repeatLimitedString.
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre><strong>Input:</strong> s = "aababab", repeatLimit = 2
+<strong>Output:</strong> "bbabaa"
+<strong>Explanation:</strong> We use only some of the characters from s to construct the repeatLimitedString "bbabaa". 
+The letter 'a' appears at most 2 times in a row.
+The letter 'b' appears at most 2 times in a row.
+Hence, no letter appears more than repeatLimit times in a row and the string is a valid repeatLimitedString.
+The string is the lexicographically largest repeatLimitedString possible so we return "bbabaa".
+Note that the string "bbabaaa" is lexicographically larger but the letter 'a' appears more than 2 times in a row, so it is not a valid repeatLimitedString.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= repeatLimit &lt;= s.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>s</code> consists of lowercase English letters.</li>
+</ul>
+
+
+**Similar Questions**:
+* [Rearrange String k Distance Apart (Hard)](https://leetcode.com/problems/rearrange-string-k-distance-apart/)
+
+## Solution 1. Greedy + Counting
+
+Store frequency of characters in `int cnt[26]`.
+
+We pick characters in batches. In each batch, we pick the first character from `z` to `a` whose `cnt` is positive with the following caveats:
+1. If the current character is the same as the one used in the previous batch, we need to skip it.
+2. On top of case 1, if the `cnt` of the character used in the previous batch is positive, then we can only fill a single character in this batch.
+
+```cpp
+// OJ: https://leetcode.com/problems/construct-string-with-repeat-limit/
+// Author: github.com/lzl124631x
+// Time: O(N)
+// Space: O(1)
+class Solution {
+public:
+    string repeatLimitedString(string s, int limit) {
+        int cnt[26] = {};
+        string ans;
+        for (char c : s) cnt[c - 'a']++;
+        while (true) {
+            int i = 25;
+            bool onlyOne = false;
+            for (; i >= 0; --i) {
+                if (ans.size() && i == ans.back() - 'a' && cnt[i]) { // the character of our last batch still has some count left, so we only insert a single character in this batch
+                    onlyOne = true;
+                    continue;
+                }
+                if (cnt[i]) break; // found a character with positive count, fill with this character
+            }
+            if (i == -1) break; // no more characters to fill, break;
+            int fill = onlyOne ? 1 : min(cnt[i], limit);
+            cnt[i] -= fill;
+            while (fill--) ans += 'a' + i;
+        }
+        return ans;
+    }
+};
+```
+## Discuss
+
+https://leetcode.com/problems/construct-string-with-repeat-limit/discuss/1784718
\ No newline at end of file